\(n_P=\frac{6,9}{31}=\frac{69}{310}(mol)\)
\(n_{O_2}=\frac{9,74}{22,}=\frac{487}{1120}(mol)\)
a) \(4P+5O_2\overset{t^o}{\rightarrow}2P_2O_5\)
\(\frac{69}{310.4}<\frac{487}{1120.5}\)
Vậy \(O_2\) dư
\(m_{O_2 dư}=\frac{487}{1120}.32-\frac{69}{310}.\frac{5}{4}.32=5,01(g)\)
b) \(m_{P_2O_5}=\frac{69}{310}.\frac{1}{2}.142=15,8(g)\)